## Quantum Country exercise 10

This is one in a series of posts working through the exercises in the Quantum Country online introduction to quantum computing and related topics. The exercises in the original document are not numbered; I have added my own numbers for convenience in referring to them.

Exercise 10. Show that $M \vert e_k \rangle$ is the $k$th column of the matrix $M$ and that $\langle e_j \vert M \vert e_k \rangle$ is the $jk$th entry of $M$.

Answer: The product of the matrix $M$ and the vector $\vert e_k \rangle$ is a vector, with the $j$th entry of that vector being the inner product of the $j$th row of $M$ with $\vert e_k \rangle$: $\left( M \vert e_k \rangle \right)_j = \sum_l M_{jl} \vert e_k \rangle_l$

We then have $\sum_l M_{jl} \vert e_k \rangle_l = M_{jk} \vert e_k \rangle_k + \sum_{l \ne k} M_{jl} \vert e_k \rangle_l$

But by the definition of $\vert e_k \rangle$ we have $\vert e_k \rangle_k = 1$ and $\vert e_k \rangle_l = 0$ for $l \ne k$. So the above reduces to $M_{jk} \vert e_k \rangle_k + \sum_{l \ne k} M_{jl} \vert e_k \rangle_l = M_{jk} \cdot 1 + \sum_{l \ne k} M_{lj} \cdot 0 = M_{jk}$

So $\left( M \vert e_k \rangle \right)_j = M_{jk}$ for all $j$, which means that $M \vert e_k \rangle$ is the $k$th column of $M$.

We now consider the product $\langle e_j \vert M \vert e_k \rangle$ for some $j$. This is the inner product of the vector $\langle e_j \vert$ with the vector $\vert M \vert e_k \rangle$, which we have just shown is the $k$th column of $M$. We therefore have $\langle e_j \vert M \vert e_k \rangle = \sum_l \langle e_j \vert_l M_{lk} = \langle e_j \vert_j M_{jk} + \sum_{l \ne j} \langle e_j \vert_l M_{lk}$

But by the definition of $\langle e_j \vert$ we have $\langle e_j \vert_j = 1$ and $\vert e_j \rangle_l = 0$ for $l \ne j$. So the above reduces to $\langle e_j \vert_j M_{jk} + \sum_{l \ne j} \langle e_j \vert_l M_{lk} = 1 \cdot M_{jk} + \sum_{l \ne j} 0 \cdot M_{lk} = M_{jk}$

We thus have $\langle e_j \vert M \vert e_k \rangle = M_{jk}$.

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